Prove the Pythagorean theorem

Erect a square outward on each of the three sides. Drop the perpendicular from $C$ to the hypotenuse and carry it straight on across the square built on $AB$. It cuts that square into two rectangles.

The claim is that each rectangle has exactly the area of one of the leg squares.

The construction

Call the square on $CA$ by the name $ACHK$, and the square on $AB$ by the name $ABDE$. Let the perpendicular from $C$ meet $AB$ at $F$ and the opposite side $ED$ at $G$.

1. Triangles $KAB$ and $CAE$ are congruent. $KA=CA$ and $AB=AE$ because they are sides of squares, and the angles between those pairs are equal — each is $\angle CAB$ with a right angle added to it. That is side–angle–side.
2. Triangle $KAB$ is half the square $ACHK$: same base $KA$, same height $AC$.
3. Triangle $CAE$ is half the rectangle $AFGE$: same base $AE$, same height $AF$.
4. Halves of equals are equal, so the square $ACHK$ equals the rectangle $AFGE$.

The mirror-image argument on the other leg gives the square on $CB$ equal to the rectangle $FBDG$. The two rectangles together are the whole square on $AB$, so
$$b^2+a^2=c^2.$$

What it costs

No ratios, no similar triangles, no multiplying of two lengths — Euclid has no real numbers to multiply with. The engine is a single fact: a triangle sheared along a line parallel to its base keeps its area. That fact is Elements I.35, and I.35 needs the parallel postulate.

That is the honest price. Every proof on this page pays it in one currency or another; this one at least makes you look at the receipt.

Drop the altitude from the right angle $C$ onto the hypotenuse, meeting it at $F$. Write $p=AF$ and $q=FB$, so that $p+q=c$.

Each small triangle shares an acute angle with the original and has a right angle of its own, so all three triangles are similar:
$$△AFC\sim △ACB\text{and}△CFB\sim △ACB.$$

From the first, $\frac{AF}{AC}=\frac{AC}{AB}$, which is $b^2=pc$.

From the second, $\frac{FB}{CB}=\frac{CB}{AB}$, which is $a^2=qc$.

Add them:
$$a^2+b^2=c(p+q)=c\cdot c=c^2.$$

Why this is the one that persuades me

It explains where the squares come from. Nothing is cut up, nothing is slid around, nothing has to be checked for fit. The two leg-squares are not mysterious areas that happen to add up — they are literally the two pieces the altitude carves out of the hypotenuse, each stretched by the same factor $c$.

Underneath it is a fact about right angles specifically: a right triangle is the only triangle that its own altitude cuts into two smaller copies of itself. The theorem is that fact, written down.

Cost: the full theory of similar triangles, which is a good deal more machinery than I.47 uses. Euclid puts similarity in Book VI for a reason. If you want the theorem early, you take the long way round.

Take four copies of the triangle and set them inside a square of side $a+b$, each turned a quarter turn from the one before. What is left in the middle is a square of side $c$.

Now count the big square's area twice.

Once, as a square of side $a+b$:
$$(a+b{)}^2=a^2+2ab+b^2.$$

Once, as its parts — four triangles and the hole in the middle:
$$4\cdot \frac{1}{2}ab+c^2=2ab+c^2.$$

Set the two counts equal and cancel $2ab$:
$$a^2+b^2=c^2.$$

The step everyone skips

That the middle piece is a square is not free. Its four sides are each $c$ by construction, but a rhombus is not a square. You also need the corners to be right angles, and they are: at each corner of the inner figure the two acute angles of a triangle lie beside the inner angle along a straight edge, so
$$\alpha +\theta +\beta =180^{\circ },\alpha +\beta =90^{\circ }\Rightarrow \theta =90^{\circ }.$$

A dissection proof is worth exactly as much as its argument that the pieces fit. Plenty of pretty dissections in circulation never make that argument and are wrong. This one earns it in a line, which is why it has lasted since the twelfth century.

Published by James A. Garfield in the New England Journal of Education, five years before he became president. It needs half the picture the four-triangle dissection needs.

Set two copies of the triangle so that the legs $b$ and $a$ lie end to end along one straight edge and the two hypotenuses meet at the point between them. The outline is a trapezoid: parallel sides $a$ and $b$, the distance between them $a+b$.

Area as a trapezoid:
$$\frac{1}{2}(a+b)(a+b)=\frac{1}{2}\left(a^2+2ab+b^2\right).$$

Area as three triangles — the two copies, plus the one spanning the two hypotenuses:
$$2\cdot \frac{1}{2}ab+\frac{1}{2}{c}^2=ab+\frac{1}{2}{c}^2.$$

Equate, double both sides, cancel $2ab$:
$$a^2+b^2=c^2.$$

The middle triangle is right-angled for exactly the reason Bhaskara's inner square is: the three angles meeting on that straight edge are $\alpha$, $\theta$, $\beta$, and $\alpha +\beta =90^{\circ }$ leaves $\theta =90^{\circ }$.

So it is the same accounting as the dissection proof, folded down the middle. Whether folding it makes the argument more elegant or merely cheaper is the kind of question this board exists to settle.